Lecture 20

Here’s a slightly better idea:

More precisely, this definition is trying to say that for any positive distance, this sequence comes within this distance of $x$.

In symbols, this can be written: $\forall\epsilon>0,\quad \exists n\in\mathbb{N}\quad\mathrm{s.t.}\quad\left|a_n-x\right|<\epsilon.$ Before we understand why this is wrong, we ought to make sure we know what this is supposed to mean.

I like to think of it as an argument with a very dangerous and unpleasant evil opponent. The evil opponent gets to choose a (positive real) distance, and we win if the sequence gets within that distance of $x$, and we lose if it doesn’t.

In order to be sure of winning, we have to know how to beat the evil opponent whatever they say.

So, in investigating how close the sequence $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots,$ gets to $1000$, then if the evil opponent is stupid enough to ask “does the sequence get within distance $100000$?” we’ll win. But we can’t just hope they’ll ask that. Instead they might ask “does the sequence get within distance $0.001$ of $1000$?”, and then we’d lose, because it never gets anywhere near there.

On the other hand, if we’re investigating how close the sequence gets to $\pi$, we win no matter what they say. If they ask “does the sequence get within distance $1000000$ of $\pi$?”, then we say “yes, $3$ is within $1000000$ of $\pi$”, and win. If they ask “does the sequence get within distance $0.0001$ of $\pi$, then we say “yes, $3.14159$ is within $0.0001$ of $\pi$”, and win. If they ask a question with an even tinier positive number in, we just take more digits and use that and say “yes”. We’re happy: the evil opponent can’t beat us.

This is a much better concept. But it’s still not right.

Here’s an example of what could go wrong. Consider the sequence $1.1,\quad 2.01,\quad 1.001,\quad 2.0001,\quad 1.00001,\quad 2.000001,\quad \ldots$ Does it converge to $1$? Does it converge to $2$?

It’s lousy as an address for either: if we give these instructions to the numberline’s village postal worker, they’ll get very annoyed as they keep walking from somewhere near $1$ to somewhere near $2$ and back again.

But according to the definition above it would “converge” to both, because it gets as close as you like to $1$ and it also gets as close as you like to $2$.

So we need to find some way of saying that it has to make its mind up eventually. The obvious thing to do is to say is that (for any $\epsilon>0$) it has to get within $\epsilon$ of $x$, and then stay within $\epsilon$ of $x$ forever.

This leads us to our final definition:
Definition: Let $x$ be a real number. A sequence of real numbers $a_0, a_1, a_2,\ldots$ is said to converge to $x$ if we have $\forall \epsilon>0,\quad \exists N\in\mathbb{N}\quad\mathrm{s.t.}\quad\forall n>N,\quad \left|a_n-x\right|<\epsilon.$

So that says “no matter what positive real $\epsilon$ our evil opponent gives us, we can point out some $N$, such that all the terms $a_{N+1}, a_{N+2}, a_{N+3}, \ldots$ are all within $\epsilon$ of $x$”.

That does an excellent job of making precise the concept of “gets close and stays close forever”, and it’s the right definition!

Now, suppose we ask whether the sequence $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots$ converges to $\pi$. It does, because no matter what $\epsilon$ our evil opponent asks about, we can find some term of the sequence beyond which all terms are within $\epsilon$ of $\pi$. For example, all terms after the $(N+1)$st term are within $10^{-N}$ of $\pi$.

Does that converge to $1000$? No, it never comes within $1$ of $1000$ (for example), so it certainly doesn’t stay within $1$ of $1000$ forever.

What about the sequence $a_0 = 1.1,\quad a_1 = 2.01,\quad a_2 = 1.001,\quad a_3 = 2.0001,\quad \ldots?$ Does that converge to anything?

No, it doesn’t. In particular, it doesn’t converge to $1$, because while it’s sometimes close to $1$, it’s also sometimes close to $2$. So there is no $N$ where $a_n$ is always within $0.1$ of $1$ for all $n>N$: all the odd-numbered $a_n$ aren’t in that range.

Similarly, it doesn’t converge to $2$, because while it’s sometimes close to $2$, it’s sometimes close to $1$. So there is no $N$ where $a_n$ is always within $0.1$ of $2$ for all $n>N$: all the even-numbered $a_n$ aren’t in that range.

So, given the difficulties we’ve had in finding the right definition, perhaps you’ll have some sympathy for the fact that it took about two centuries to sort real analysis out properly. In what remains of the course I’ll try to make you like this definition.

Convergence proofs

Now we get back to the subject of convergence.

We say that a sequence $(a_i)_{i\in\mathbb{N}} = a_0, a_1, \ldots$ is convergent if it converges to some $x$.

Here’s a very important fact (which is only true because of all that work we put in finding a good definition):

Proposition

A sequence $a_0, a_1, \ldots$ cannot converge to two different real numbers $x$ and $y$.

Before we prove this, let’s think about what a proof might look like. Here’s a picture of a numberline, to help show us what would happen if a sequence was convergent to $x$ and to $y$:

The two bars at the bottom of that diagram were deliberately chosen not to overlap (I made each of them extend one third of the distance from $x$ to $y$, leaving another third of the distance between them in the middle).

And since it converges to $x$ and $y$, eventually all the terms of the sequence should be within the interval around $x$, and also all of them within the interval around $y$, which is a contradiction since the intervals don’t meet.

Let’s do the working carefully.

Proof

We’ll prove this by contradiction. So, suppose it can: suppose that there is a sequence $a_0, a_1, \ldots$, which converges to two different real numbers $x$ and $y$. Without loss of generality, we may take $x<y$.

Since the sequence $a_0,a_1,\ldots$ converges to $x$, there is some $N$ such that, for all $n>N$, we have $\left|a_n - x\right|<\frac{y-x}{3}$.

Since the sequence $a_0,a_1,\ldots$ converges to $y$, there is some $M$ such that, for all $n>M$, we have $\left|a_n - y\right|<\frac{y-x}{3}$.

But then, using the triangle inequality, for any $n$ bigger than both $N$ and $M$, we have $y-x=\left|y-x\right| {}\leq\left|y-a_n\right|+\left|x-a_n\right| {}<\frac{y-x}{3}+\frac{y-x}{3} {}=\frac{2}{3}(y-x)$ which is a contradiction as $y-x$ is positive.

As a result, if a sequence is convergent, there is a unique real number to which it converges; we call that the limit of the sequence.

I understand you will have met the subject of convergence in MAS110.

That course is about streetfighting, and there you’re encouraged to use any techniques you have to hand without worrying too much about what it means.

This is a course about proof in mathematics, and tracing reasoning back to basic principles. If I set problems about convergence in MAS114, I need you to give a rigorous proof, with everything traced back to the definition of convergence (unless you’re told otherwise), rather than using the slightly vaguer methods and extra theorems you saw there!